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Bad White [126]
2 years ago
6

Select the three expressions that are equivalent to 2 (4 - 3x) + (5x-2) ​

Mathematics
2 answers:
Anni [7]2 years ago
8 0

Answer:

Step-by-step explanation:

2 (4 - 3x) + (5x-2)  can be solved in several basic steps.  First, distribute multiplication in 2(4 - 3x), obtaining 8 - 6x and (overall) 8 - 6x + 5x - 2.

Combine like terms:  -x = -10, or x = 10

Ivanshal [37]2 years ago
4 0

Answer:

The answer correct the questions is 11x-10

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Solve. −1.2z − 8.2 > −9.7
leonid [27]
<span>−1.2z − 8.2 > −9.7</span>
−1.2z > - 1.5
  z < 1.25


3 0
2 years ago
A payment based on actual working time or actual production
AleksandrR [38]
I believe this is called a wage
6 0
3 years ago
If a procedure meets all of the conditions of a binomial distribution except the number of trials is not​ fixed, then the geomet
morpeh [17]

Answer:

The probability is 0.0428

Step-by-step explanation:

First, let's remember that the binomial distribution is given by the formula:

P(X=k) =\left[\begin{array}{ccc}n\\k\end{array}\right] p^{k}(1-p)^{n-k} where k is the number of successes in n trials and p is the probability of success.

However, the problem tells us that when there isn't a number of trials fixed, we can use the geometric distribution and the formula for getting the first success on the xth trial becomes:

P(X=x) = p(1-p)^{x-1}\\

The problem asks us to find the probability of the first success on the 4th trial (given that the first subject to be a universal blood donor will be the fourth person selected)

Using this formula with the parameters given, we have:

p = 0.05

x = 4

Substituting these parameters in the formula and solving it, we get:

P(X=4) = 0.05(1-0.05)^{4-1}\\P(X=4) = 0.05 (0.95)^{3}\\P(X=4) = 0.05(.8573)\\P(X=4) = 0.0428

Therefore, the probability that the first subject to be a universal blood donor is the fourth person selected is 0.0428 or 4.28%

7 0
3 years ago
Show that 1n^ 3 + 2n + 3n ^2 is divisible by 2 and 3 for all positive integers n.
Dmitry_Shevchenko [17]

Prove:

Using mathemetical induction:

P(n) = n^{3}+2n+3n^{2}

for n=1

P(n)  = 1^{3}+2(1)+3(1)^{2} = 6

It is divisible by 2 and 3

Now, for n=k, n > 0

P(k) = k^{3}+2k+3k^{2}

Assuming P(k) is divisible by 2 and 3:

Now, for n=k+1:

P(k+1) = (k+1)^{3}+2(k+1)+3(k+1)^{2}

P(k+1) = k^{3}+3k^{2}+3k+1+2k+2+3k^{2}+6k+3

P(k+1) = P(k)+3(k^{2}+3k+2)

Since, we assumed that P(k) is divisible by 2 and 3, therefore, P(k+1) is also

divisible by 2 and 3.

Hence, by mathematical induction, P(n) = n^{3}+2n+3n^{2} is divisible by 2 and 3 for all positive integer n.

3 0
3 years ago
Add (4x^2-xy-y^2) and (x^2+5xy+8y^2) simplify your answer:
bazaltina [42]

Answer:

To find a, b, and c, rewrite in the standard form ax2+bx+c=0ax2+bx+c=0.

a=1, b=3, c=0

8 0
3 years ago
Read 2 more answers
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